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Let $P$ be an integer valued polynomial with degree $> 1$, is it a theorem that the values $P(n)$ have arbitrarily large prime factors?

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Yes suppose that all values of $P(n)$ are divisible by the prime numbers $p_1,p_2,...,p_r$ only. Now consider $P(P(0)p_1p_2...p_r)$, if $P(0)$ is different from zero we find that none of the primes $p_1,p_2,...,p_r$ divides $P(P(0)p_1p_2...p_r)/P(0)$.(a contradiction). If $P(0)=0$, then $P(0)$ is divisible by arbitrary large primes.

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Here is an interesting alternate approach to this problem. –  Dan Brumleve Dec 8 '12 at 0:18
    
Thank you for the nice link –  Amr Dec 8 '12 at 0:20
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Dear @Amr: I am not sure if your statement "none of the primes $p_1, \dots, p_r$ divides $P(P(0)p_1\dots p_r)$" is true unless you divide by $P(0)$ when $P(0) \neq 0$. –  Rankeya Dec 8 '12 at 1:15
    
Yes you are right. I will edit my answer –  Amr Dec 8 '12 at 9:07
    
why /P(0) and how do you know p_1 doesn't divide that? –  user51427 Dec 11 '12 at 4:24
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